Two moles of helium gas are taken over cycle ABCDA , as shown in P-T diagram. net work done on gas i

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11.Thermodynamics

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Two moles of helium gas are taken over the cycle $ABCDA$, as shown in the $P-T$ diagram. The net work done on the gas in the cycle $ABCDA$ is ...... $R$

A

$279$

B

$1076 $

C

$1904$

D

$0$

(AIEEE-2009)

Solution

The net work in the cycle $ABCDA$ is

$W = {W_{AB}} + {W_{BC}} + {W_{CD}} + {W_{DA}}$

$ = 400R + 2.303nRT\log \frac{{{P_B}}}{{{P_C}}} + \left( { – 400R} \right) – 414R$

$ = 2.303 \times 2R \times 500\log \frac{{2 \times {{10}^5}}}{{1 \times {{10}^5}}} – 414R$

$ = 693.2\,R – 414\,R = 279.2\,R$

Standard 11

Physics

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